Here's a thought experiment: You have a bar(log) made of a solid uniform material of low thermal conductivity (say wood) and placed one end close to a source of heat (say hot oil). Will the other end of the log ever get to the same temperature as the end near the heat source? Let's take ideal conditions so this happens in a vacuum. Now obviously the result of this problem may well boil down to numbers so here are a few: \(k_{wood} = 0.2 W\left(m\cdot K \right)^{-1}\) \(T_{oil} = 300^{o}C\) Temp of log to start with = \(20^{o}C\) Length of log= 10m Radius of log = 0.1m If it does eventually reach that temperature, is there a length of wood where no heat energy at all from the oil will ever reach the other end?
You've got one end near a high-temperature source of heat. What happens next depends on what's at the other end. If you keep the other end cold, then the other end won't ever heat up. But if there's no way for the other end to lose energy, then it will eventually heat up until the whole log is at the same temperature.
In a closed system the entire closed environment will eventually reach equilibrium and so the entire log will reach the same temperature. Obviously to maintain the constant temperature of the oil, energy must be added to the system and so it is not closed. In an open system the efficiency of thermal radiation is affected by the distance and so the only way that the entire log can reach the same temperature is if the heat conductivity through the log itself can more than offset the decline in thermal heat transfer efficiency at the near end vs. the far end. The answer depends on the relationship between the conductivity of the log, vs. the efficiency of thermal heat transfer across that same distance. In an open system there is ongoing heat loss from the log. If the length of the log allows more heat loss than the conductivity of the log permits heat conduction, then the far end will never reach the temperature of the near end, IHMO.
Solve the heat equation for a cylinder or cuboid or rectangle where one of the ends is kept at a constant temperature (which would be your boundary condition).
If it is in vacuum then you don't need to worry about heat loss by any means except radiation. I guess you could assume there was no incoming radiation if you keep it in a giant dark vacuum room (maybe deep space). The heat will reach the other end of the log faster than you will lose it to radiation, at least at first, since the cold end of the log will lose practically no energy via radiation. (radiation heat loss goes with T^4 while conduction is essentially linear in T, so while the cold end of the log is still cold energy will reach it by conduction more efficiently than it can be radiated away. Actually, your starting temp of 20 C will be deciding factor, if it is cold enough what I said is true, otherwise the "cold" end of the log will cool down even more until the radiated energy is equal to the amount the makes it through the log.)
Exactly, that was the point. That's what i am assuming Interesting, so does this mean that the far end will at some stage reach the same temperature at a point very close to the hot oil? So what are you saying then? Does the 20 C far end cool down or will it reach a temperature cose to 300 C?
Well there is no active refrigeration at the other end if that's what you mean by keeping it cold Please Register or Log in to view the hidden image! Well that's part of the problem isn't it? You gotta try to figure out what the ways of losing energy are. And if it will affect the temperature of the far end significantly enough.
No, we are assuming open. Let's say (as Kurros suggested) deep space. Can you explain this a bit more? By keeping one end close to 300 C, is it possible to eventually have the other side at exactly the same temp? Even with radiation occuring in between? Or must it always be a little cooler?
Actually I just meant explain the last sentence (not the open and closed system bit). i.e. What did you mean by this exactly?
I still haven't bother to put the numbers in so I don't know exactly, but in the situation we have agreed on the final result will be a log at 300 C (or whatever you said the oil temp was) and some cooler temperature at the other end, with some temperature gradient in between. I think the gradient would be more complicated than just linear. Exactly what temperature the cool end of the log ends up as depends on the numbers. But the essence of it is that a balance between energy lost by radiation and energy gained from conduction from the oil heat bath will be reached.
OK, we are getting somewhere. So you are pretty sure that the other end of the log will never be the same temperature as the hot end (assuming an open system). The temperature gradient is also what I intuitively thought. Now it begs the question that is there ever going to be a point on a very long log where no heat is reached a all and it remains at the environmental temperature?
I think yes. First, in this scenario there really isn't much concept of environmental temperature, being a perfect vacuum. Perhaps you could say the CMB is there, then the environment is at 3 K or whatever it is, so lets go with that. This kind of radiation temperature is fairly different from the log being in some gaseous medium at 3K though. Anyway, here is my vague reasoning. Say we have some infinitesimally thin slice of this log. The change in temperature of this part of the log is some function of the change in heat energy of the slice, I think just linear but it doesn't matter. I figure a vague equation describing the change in heat energy of this slice is: \( \Delta Q = K_C \Delta T - K_{R}(T)^4 \) where \( \Delta Q \) is the change in heat energy of the slice, \( K_C \) and \( K_{R} \) are radiative and conductive constants which depend on the wood etc, \( \Delta T \) is the temperature gradient of the slice and \(T\) is the average temperature of the slice. Now if we go to a really long log then the temperature gradient in the slice goes to zero far from the heat source, so conduction is suppressed. At this point if the T is anything other than zero K then heat is lost (or if there is some environmental temperature until that temperature is reached). Basically a long log isolates the far end from the heat source, no matter how awesome a conductor of heat it is. Of course how long your log needs to be to achieve this condition depends a lot on how good a conductor it is. Admittedly my equation above is kind of dodgy, not in the least because it doesn't take area into account, and an infinitesimal slice can't radiate out of the log because the exposed surface area is zero, but I think the logic more or less works. If we assume it is a bigger chunk, not so infinitesmal, then my argument is a bit more solid.
Thanks Kurros, I agree. Something else just struck me as I read your post. The better a conductor the material is, the better potential heat has to travel along the log, however in actuality, that distance is largely reduced because the better a conductor the material is, the better heat will be conducted out from the interior of the log to replace heat lost by radiation at the surface Please Register or Log in to view the hidden image! This adds an extra dimension to the problem and so there is a limit to how good a conductor you should use based on the thickness of the log.
Haha, yes that is a factor I guess. Although clearly you should use a log that is a good conductor on the inside and a good insulator on the outside Please Register or Log in to view the hidden image!. Although actually it depends how opaque the material is to thermal radiation, if it is pretty transparent then the heat can radiate away straight from the inside, through the outer layers of the material. You need the outside to reflect the radiation back inwards Please Register or Log in to view the hidden image!.
True, although in the first post I mentioned that it should be a uniform material (meaning density etc too). So that should be no conductance differences between surface and interior. But yes, if the material were glass or ice then the entire section of the log would radiate Please Register or Log in to view the hidden image!
